Question

Find the equation of the circle with the center $(\dfrac{1}{2},\dfrac{1}{4})$ and radius $\dfrac{1}{12}$

Answer 1

The equation of a circle with center (h, k) and radius r is given as $(x- h)^2 + (y - k)^2 = r^2$

Given that the center $(h, k) = (\dfrac{1}{2}, \dfrac{1}{4})$and radius $(r) =\dfrac{1}{12}$.

Therefore, the equation of the circle is

$(x -\dfrac{1}{2})^2+ (y -\dfrac{1}{4})^2= (\dfrac{1}{12})^2$

$⇒x^2+y^2-2.x.\dfrac{1}{2}-2.y.\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{16}=\dfrac{1}{144}$

$⇒x^2+y^2-2.x.\dfrac{1}{2}-2.y.\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{144}=0$

$⇒144x^2-144x+144y^2+36-72y+9-1=0$

$⇒36x^2+36y^2-36x-18y+11=0$ (Ans.)